#### Exercises - Uniformly Accelerated Motion

**Exercise 1:**

A climate activist stuck himself on the street in the city center. A car approaching at 50 km/h brakes immediately. What braking distance does the car need if a reaction time of one second has to be taken into account? The braking deceleration (negative acceleration) is -8 m/s^{2}.

The reaction distance s_{react} represents a uniform, straight-line movement:

s_{react} = v_{0} ∙ t_{react} = 50 km/h ∙ 1s = 13,89 m/s ∙ 1s = 13,89 m

( 50km/h = 50 000 m / 3600 s = 13,89 m/s )

The braking distance **s _{braking}** can be determined by a negatively accelerated movement:

To use this equation, you must know that the braking distance must be calculated at the point in time at which the velocity has reached the value 0 m/s. We started with a speed of 13.89 m/s. Thus follows:

v

^{2}- v_{0}^{2}= (0 m/s )^{2}- (13,89 m/s)^{2}= - (13,89 m/s)^{2}

The total stopping distance is the sum of the reaction distance and the braking distance:

s_{total} = s_{react} + s_{braking} = 13,89 m + 12,06 m = 25,95 m